Number of Possibilities Question Help

[thunc14]

'Nicole's favourite ice cream store offers eight different flavours. Nicole wants to buy three scoops of ice cream which are not all of the same flavour. How many possibilities does she have to do this? *The order of the flavours doesn't matter*.

I'm very used to just doing this as an '8*7*6' type of problem. Since order doesn't matter, I know that wouldn't be the case, since this is a combination problem. The whole 'not all of the same flavour' bit confuses me. So assuming two of them can be the same flavor, would this then be a combination with replacement problem...?

So would the formula to solve this then be:

(r+n-1)!/(r!(n-1)!), then subtract 8 from that total because that's the number of possibilities where there are 3 ice cream flavors?

For what it's worth, this isn't pulled from a permutation/combination class, it's from an old international math competition paper.

Many thanks in advance!

 

[JeffM]

You are thinking well.

Suppose the problem said, "How many ways can she pick three flavors all different?"

Suppose the problem said "How many ways can she pick two identical flavors and a third different flavor?"

How do those relate to your actual problem?

 

[Dr.Peterson]

So would the formula to solve this then be:

(r+n-1)!/(r!(n-1)!), then subtract 8 from that total because that's the number of possibilities where there are 3 ice cream flavors?

Yes, though you haven't said what r and n are, and I think you meant "where there are 3 of the same flavor".

 

[pka]

'Nicole's favourite ice cream store offers eight different flavours. Nicole wants to buy three scoops of ice cream which are not all of the same flavour. How many possibilities does she have to do this? *The order of the flavours doesn't matter*.
So would the formula to solve this then be:
(r+n-1)!/(r!(n-1)!), then subtract 8 from that total because that's the number of possibilities where there are 3 ice cream flavors?

Yes that is the rule. To place [imath]K[/imath] identical objects into [imath]N[/imath] distinct cells can be done in [imath]\dfrac{(N+K-1)!}{N-1)!(K!)}[/imath] ways.
In this question the acts of choosing are identical and the flavors are the distinct cells.